Question

3а2+а-4 разложиье на множители плиз

Answer 1

3a^2 + a - 4 = (a - a1)(a - a2) = (a - 1)(a -(-4/3)) = (a - 1)(a + 4/3)
D = b^2 - 4ac = 1 + 4(-4)3 = 1 + 48 = 49. $sqrt{D} = sqrt{49} = 7$.
a1 = (-b + $sqrt{D}$ )/2a = (-1 + 7)/6 = 1.
a2 = (-b - $sqrt{D}$ )/2a = (-1 - 7)/6 = -4/3.